Skip to main content

Full text of "Mathematical And Physical Papers - Iii"

See other formats

The equation (17) may be reduced in a similar manner, and we get finally
which is an exact differential by virtue of (22).
12. Passing to polar co-ordinates, let r be the radius vector drawn from the origin, 6 the angle which r makes with the axis of x, and let R be the velocity along the radius vector, © the velocity perpendicular to the radius vector : then
x = r cos 6, is = r sin 6, u = R cos 6 — © sin 6, q = R sin 6 -f- © cos 0. Making these substitutions in (20), (22), (23), and (25), we obtain r sin 6 (RrdQ - ®dr) = cfyr .............. ....(26),
sin0 d (   1   cfyA +~~"
r sin d \dtdr         r dtdd
We must now determine ^ and ty2 by means of (27) and (28), combined with the equations of condition. When these functions are known, p will be obtained by integrating the exact differential which forms the right-hand member of (29), and the velocities R, ©, if required, will be got by differentiation, as indicated by equation (26). Formulae deduced from (4) and (5) will then make known the pressure of the fluid on the sphere.
13. Let | be the abscissa of the centre of the sphere at any instant. The conditions to be satisfied at the surface of the sphere are that when r = rv the radius vector of the surface, we have
Now rt differs from a by a small quantity of the first order, and since this value of r has to be substituted in functions which are already small quantities of that order, it will be sufficient to put r = a. Hence, expressing R and © in terms of -v/r, we get
>, ^ = a2sin0cos<9^,  when r- a ..... (30).